Let $f(x)=\dfrac{1}{x^3}-\dfrac{1}{x}+x^2$. $f'(x)=$
The strategy We can first rewrite each rational term of $f$ as a negative power of $x$. Then, the derivatives of these terms can be found using the power rule : $\dfrac{d}{dx}(x^n)=n\cdot x^{n-1}$ (Remember that this applies even when $n$ is negative.) Rewriting rational terms as negative powers $\begin{aligned} f(x)&=\dfrac{1}{x^3}-\dfrac{1}{x}+x^2 \\\\ &=x^{-3}-x^{-1}+x^2 \end{aligned}$ Differentiating using the power rule $\begin{aligned} &\phantom{=}f'(x) \\\\ &=\dfrac{d}{dx}(x^{-3}-x^{-1}+x^2) \\\\ &=\dfrac{d}{dx}(x^{-3})-\dfrac{d}{dx}(x^{-1})+\dfrac{d}{dx}(x^2) \\\\ &=-3x^{-4}-(-1)x^{-2}+2x \\\\ &=-\dfrac{3}{x^4}+\dfrac{1}{x^2}+2x \end{aligned}$ In conclusion, $f'(x)=-\dfrac{3}{x^4}+\dfrac{1}{x^2}+2x$.